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3   SPECTRUM OF A COMMUTATIVE BANACH ALGEBRA       321

(15.3.6)    Suppose that there exists an element XQ e A which, together with e,
generates a dense subalgebra of A. (in other words, as P runs through the set
C[X] of polynomials in X with complex coefficients, the subalgebra consisting
of the P(x0) is dense in A). Then the mapping % H+ x(xo) is a homeomorphism of
X(A) onto SpA(^0).

Notice first that the elements P(x0) for which the coefficients of the poly-
nomial P are of the form a + /?/ with a and /? rational, form a dense subset
of A, and therefore A is separable. The mapping X\-+X(XQ) is continuous
and surjective (15.3.4), and X(A) is compact and metrizable (15.3.2); hence
it is enough to show that this mapping is also injective (3.17.12). But if
XiOo) = %2(*o) for two characters xl9 x2, then also Xi(^M) = X2(P(*o))
for all polynomials P; since Xi and #2 are continuous on A, we deduce
that Xi = #2 (3.15.2), and the proof is complete.

Examples of Spectra and GeJfand Transformations

(15.3.7)    Let X be a metrizable compact space and A = #C00 the Banach
algebra of continuous complex-valued functions on X. Then the characters
of A are the Dirac measures ex (x e X) (13.1.3), and the mapping xt-*sx is a
homeomorphism of X onto the spectrum X(A). For in this situation we have
A' == MC(X), since every measure on X is bounded (13.20), and the weak
topology on A' is by definition the vague topology (13.4). A character is
therefore a measure /* ^ 0 (because ^(1) = 1). We shall show that Supp(ju)
consists of a single point. If Supp(ju) contained two points a ^ b, then a and b
would have disjoint neighborhoods U and V, respectively, hence we should
have functions/and g belonging to A, with supports contained in U and V,
respectively, and such that X/) = 0 and n(g) = 0. But then fg = 0 so that
/*(/)/*(#) = V>(f&) = 0> giving a contradiction. If Supp(ju) = {x}, then the
relation ex(/) = 0 implies X/) = 0, and therefore (A.4.15) there exists a
scalar a such that ju = aex; since ju(l) = 1, it follows that a = 1. The continuity
of the mapping xh->ex is  immediately  seen: for each /eA, we have
</, sx  exoy =f(x)  f(xo), and since/is continuous, for each 8 > 0 there
exists a neighborhood V of XQ in X such that \f(x)  f(x0)\ ^ d for all x e V.
In order to prove that :cf->^ is a homeomorphism, it is enough (12.3.6) to
show that this mapping is injective; and this is clear, because exi and eX2 have
distinct supports if x1 ^ x2 . We have (^/)(ej = ex(f) =f(x) for all x e X
and all/e A, and therefore when we identify X and X(A) by means of the
mapping x\-^sx.> the Gelfand transformation becomes the identity mapping.

(15.3.8)    Let D be the closed unit disk |C| ^ 1 in C, and let A = ^(D) be the
closed subalgebra of ^C(D) consisting of the continuous functions on D whichSp(#).n particular, the image of C under the